Problem: The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives $27$ years; the standard deviation is $2.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living longer than $31.6$ years.
Answer: $27$ $24.7$ $29.3$ $22.4$ $31.6$ $20.1$ $33.9$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $27$ years. We know the standard deviation is $2.3$ years, so one standard deviation below the mean is $24.7$ years and one standard deviation above the mean is $29.3$ years. Two standard deviations below the mean is $22.4$ years and two standard deviations above the mean is $31.6$ years. Three standard deviations below the mean is $20.1$ years and three standard deviations above the mean is $33.9$ years. We are interested in the probability of a zebra living longer than $31.6$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the zebras will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the zebras will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $22.4$ years and the other half $({2.5\%})$ will live longer than $31.6$ years. The probability of a particular zebra living longer than $31.6$ years is ${2.5\%}$.